5 Answers
5

The previous answers assume that $A$ and $B$ are square. This is fair because it is not generally possible for $A\otimes B$ to be invertible when $A$ and $B$ are rectangular.

In the more general setting where $A$ is $m\times n$ and $B$ is $r\times s$ (and $mr = ns$) it is still easy to see that $A\otimes B$ is equivalent to $B \otimes A$ up to permuting the rows and columns.

Both $A \otimes B$ and $B \otimes A$ contain exactly the same entries $a_{ij} b_{kl}$. More explicitly, one places $a_{ij} b_{kl}$ in row $(i-1)r+k$ and column $(j-1)s+l$, and the other places it in row $(k-1)m+i$ and column $(l-1)n+j$. Note that the row numbers are uniquely determined by $i,k$ and the column numbers determined independently by $j,l$.

Do you know that if two matrices $A$ and $B$ represent two linear maps $a : V \to V'$ and $b : W \to W'$ (with respect to some chosen bases of the vector spaces $V$, $V'$, $W$ and $W'$), then the Kronecker product $A \otimes B$ represents the linear map $a \otimes b : V \otimes W \to V' \otimes W'$ (with respect to the appropriately defined bases: e.g., if our bases of $V$ and $W$ are $\left(e_i\right)_{i \in I}$ and $\left(f_j\right)_{j \in J}$, then for $V \otimes W$ we use the basis $\left(e_i \otimes f_j\right)_{\left(i,j\right)\in I\times J}$) ? If so, then you will realize that your question is equivalent to the following question: If the tensor product $a \otimes b$ of two linear maps $a$ and $b$ is invertible, then so is $b \otimes a$. And this follows from the following commutative diagram:

"if and only if" is wrong, as $A$ could be a $0\times 0$-matrix, thus rendering $A \otimes B$ invertible for trivial reasons no matter what $B$ is. This argument requires a more thorough handling of border cases.
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darij grinbergJan 30 at 18:12

@darij: Thank you, I edited to make clear that my answer only works under additional assumptions.
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Jonas MeyerJan 30 at 19:06

Case 1. If $p=q,m=n$, that is, if $A,B$ are square matrices, then $A\bigotimes B$ and $B\bigotimes A$ are similar.

Case 2. Otherwise it is false. Yet $A\bigotimes B$ and $B\bigotimes A$ have same singular values $(\sigma_i\tau_j)_{i,j}$ where $(\sigma_i)_i, (\tau_j)_j$ are the singular values of $A,B$. Indeed, if one has the SVD $A=U_1D_1V_1,B=U_2D_2V_2$, then we deduce the SVD $A\bigotimes B=(U_1\bigotimes U_2)(D_1\bigotimes D_2)(V_1\bigotimes V_2)$ and $B\bigotimes A=(U_2\bigotimes U_1)(D_2\bigotimes D_1)(V_2\bigotimes V_1)$.

A direct consequence is the following generalization of the result requested by the OP: $rank(A\bigotimes B)=rank(B\bigotimes A)=rank(A)rank(B)$.